By RBarryYoung


2015-03-12 21:07:07 8 Comments

I am not a physicist (math/comp-sci) but I understand that Angular Momentum is supposed to be conserved. I find this confusing because there seems to be many simple, common cases where a restrained, revolving body, when released seems to lose it's angular momentum.

For example, let's say that a hammer thrower is revolving a 10kg hammer around herself at a 1 meter distance from their combined center of mass every second. Taking the hammer as a point-mass, then before release, the hammer and the thrower each have an angular momentum of

$l\omega = (mr^2)*(2\Pi) = (10*1*1)*(2*3.14..) =~ 62.8 kg\text{-}m^2/s$

Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any.

Yes, I know that the hammer is still rotating (spinning) as it did when it was revolving around the thrower, so that angular momentum seems to be conserved, but the angular momentum from revolution (orbiting) seems to be gone. AFAIK, it's not transferred to it's spin angular momentum (observation seems to bear this out).

Nor is it transferred to the thrower. Besides having no mechanism for this (she let go of it), if it did, we would have noticed hammer throwers being knocked over by having their angular momentum suddenly doubled.

So where does it go? Or is it not actually conserved in this case?

5 comments

@user75366 2015-03-17 08:34:50

Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any.

A body does not have angular momentum wrt to a point C only when it is circling around it, you know that planets have elliptical orbits and do have L

If a body H has linear momentum p it has also and angular momentum L (you may say virtual to have an intuitive understanding) wrt any point in space. You may say that it has no L (or L= 0) when the point C lies on its trajectory, like body B in the sketch:

enter image description here

The formula is $L =mvr$ and r is the distance between H and the parallel to its trajectory passing through point C.

If the hammer is approaching point C ($H_1$) at $v = 2\pi$, is ($H_2$)circling/ orbiting around C or flies off the tangential ($H_3$)at the same speed its L will not change: $L = p*r= 10*2\pi*1= 20\pi$. If its speed changes L will change accordingly, but only if an external force acts on it. In the same way the thrower will alter her L if she stops rotating after releasing the hammer.

If you examine the sketch you will see that if you consider the actual distance D from H to C and multiply it (or the vector v) by the the sine of the angle the vector makes with the line HC (=D) the value of the angular momentum does not change

@image 2015-03-16 23:05:15

Angular momentum is conserved in this example!

As you already stated, the angular momentum of the thrower doesn't change after the hammer is released.

Consider the hammer being in rotation around the origin of our coordinate system for $t < 0$: $$ \vec{r}(t) = r_0 \ \ (cos(\omega t), sin(\omega t), 0)^T $$. Its momentum is therefore given by: $$ \vec{p}(t) = m \vec{v} = m \dot{\vec{r}} = m r_0 \omega \ \ (-sin(\omega t), cos(\omega t), 0)^T $$ Now we know that its angular momentum is given by: $$ \vec{L}(t) = \vec{r}(t) \times \vec{p}(t) = m r_0^2 \omega \ \ (0,0,1)^T $$

Assume that the hammer is released at $t=0$. It will then travel on a straight line, parallel to $\vec(p)(0)$. One can express this movement by: $$ \vec{r}\ '(t) = \vec{v}\ ' \cdot t + \vec{r}(0) = \frac{\vec{p}(0)}{m} \cdot t + \vec{r}(0) = r_0 \ \ (1, \omega t, 0)^T $$ It clearly has the momentum: $$ \vec{p}\ '(t) = \vec{p}(0) = m r_0 \omega \ \ (0, 1, 0)^T $$ By calculation one gets for the angular momentum after the release $$ \vec{L}\ '(t) = \vec{r}\ '(t) \times \vec{p}\ '(t) = m r_0^2 \omega \ \ (0,0,1)^T $$, which is the same as before the release.

In general angular momentum need not be conserved in every process. Only if the underlying action (in terms of Lagrangian formalism) is invariant under rotation around an axis, the angular momentum in direction of that axis is conserved.

@batee 2015-03-12 21:54:24

When the hammer thrower swings the hammer, according to Steiner's Theorem, the moment of inertia of the system is combined from a mass point at a distance from the center of rotation and the body rotating around its center of mass

$$I = I_c + m \times r^2$$ so our complete moment of inertia is combined from 4 parts.

$$I_{\text{complete}} = (I_{c_1} + m_1 \times r_1^2) + (I_{c_2} + m_2 \times r_2^2)$$

From this we can calculate the angular momentum.

When the hammer is released, the angular momentum is split for both objects. Both remain rotating around their center of mass, but the remaining angular momentum (from $m \times r^2$) is converted to linear momentum.

@Sofia 2015-03-12 21:59:43

angular momentum and linear momentum are two different physical quantities, with different dimensions and units. Angular momentum has the units Kg $\cdot$ m$^2$/s, and linear momentum has Kg $\cdot$ m/s.

@batee 2015-03-12 22:02:51

the conversion is based on conservation of kinetic energy

@Sofia 2015-03-12 22:05:12

These are three different physical quantities, each one with its own conservation law: energy conservation, linear momentum conservation, and angular momentum conservation.

@batee 2015-03-12 22:21:56

to prove my point: kinetic energy from rotation, Ekr = m * r^2 * omega^2 / 2 ; kinetic energy from translation, Ekt = m * v^2 / 2 ; Ekr = Ekt ; v^2 = omega^2 * r^2 ; v = omega * r ; which is according to the eulers equation; so the linear momentum: P = v * m = omega * r * m

@Sofia 2015-03-12 23:08:27

If you want to prove smth. against the conservation law, I am sorry, but the conservation laws were never violated until now.

@jaromrax 2015-03-12 21:32:06

Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate.

In fact they have (as one system) the same angular momentum all the time.

When you have a projectile that aims a bit off center towards the target, the angular momentum of the system is nonzero. I think you find it under name collision parameter, usually $b$. If collision parameter is zero, angular momentum is zero.

The hammer and thrower is just time reversed situation. Angular momentum conserved. He might dissipate some his remaining bits of energy into the ground.

To better constraint the problem: imagine that nothing else exists in universe, only the hammer and the thrower. Forget any rotations of hammer or thrower. For eternity the system of hammer+thrower will keep the total angular momentum. Once you remove your thrower from the system, it is another excersise.

Small remark: The angular momentum is not partially there and there, the complete system has it.

The same picture is in first pages of nuclear physics textbooks, particle $a$ going to nucleus $C$ a little bit off axis. The distance between center of $C$ and axis of flight is that $b$. And you have defined the angular momentum of the system $=b \cdot p$ for any situation, any force between $a$ and $C$, any time before or after. Notice, that in an empty universe, the thrower cannot properly throw a hammer starting from rest (due to the conservation).

@zeldredge 2015-03-12 21:35:09

Note that the hammer-thrower can't throw the hammer forward AND in a line with the axis of rotation, because he'd have to let go of it when it was in front of him, and at that point it's traveling purely tangentially.

@RBarryYoung 2015-03-12 23:43:29

Either way, I would appreciate seeing the mathematical derivation that demonstrates that the hammer still has an angular momentum of apprx. 62.8 ...

@jaromrax 2015-03-13 08:04:34

@RBarryYoung - The hammer does not have angular momentum - it is the system of thrower and hammer who has angular momentum. It would be incorrect to remove a part of a system and demand a transfer of a.m. to just one part of it.

@jaromrax 2015-03-13 12:16:18

@GreenRay - so, what angular momentum has the referee, who is hit directly (on axis) by this hammer to the head? If hammer itself had the angular momentum, the referee should spin then, should not? (Which is actually possible anyway :)

@jaromrax 2015-03-13 16:14:56

@GreenRay - bet on it, that referee - having the hammer in his head - has quite a momentum and velocity. However 0 angular momentum if well hit.

@BowlOfRed 2015-03-12 23:07:53

Now, after the hammer is released, the thrower still has her same angular momentum (and has to slow herself down), but the hammer no longer seems to have any.

Even though the hammer isn't rotating around the axis, it still has the same angular momentum it had at release with respect to the original axis.

So the formula $$L = mvd$$ is correct both for a point mass orbiting an axis at a given radius $d$, or for a point mass moving in a straight line, with $d$ being the distance of closest approach to the axis of consideration.

So the angular momentum is conserved, and is partially in both the hammer and the thrower.

@RBarryYoung 2015-03-12 23:44:06

You say, "..it still has the same angular momentum..". Could you demonstrate that please? That is, show how the hammer still has an angular momentum of 62.8.

@Ivan Lerner 2015-03-12 23:48:23

The momentum is conserved on the system, not its parts. The momentum of the hammer might not be the same, but the total will.

@RBarryYoung 2015-03-12 23:54:01

@IvanLerner I already demonstrated in my question how the $l\omega$ of the hammer was not distributed to the other parts. If you think that I am wrong about that, please provide some justification.

@BowlOfRed 2015-03-12 23:58:49

@RBarryYoung, your original calculation was only for the angular momentum of the hammer. Upon release, it retains its speed which is $C/t = 6.28m/s$ $L = mvd$. With $m=10kg$, $v=6.28m/s$, and $d=1m$, the product is $62.8kg m^2/s$

@Ivan Lerner 2015-03-13 00:00:31

I don't think you've showed that the angular momentum of the thrower will be the same, and in fact it might not be, it is only the same if the hammers is also the same.

@RBarryYoung 2015-03-13 00:06:42

@IvanLerner It is the same because before release the hammer and the thrower are in balance (approximately). Hammer throwing relies on this, otherwise, the thrower wouldn't be able to stay upright while hanging on to that much angular momentum. Again, if you disagree (and I could be wrong) then show me the math, or a better explanation/derivation.

@RBarryYoung 2015-03-13 00:10:25

@BowlOfRed That doesn't answer my question, what about when the hammer is 5m away? Now $d ~= 5m$, so $mvd = 314$. Or 10m, 20m or 30m away? (you can assume no friction/no gravity for this)

@Ivan Lerner 2015-03-13 00:11:23

No, my bad, it should be the same.

@BowlOfRed 2015-03-13 00:24:28

@RBarryYoung, the $d$ there is for point of closest approach between the line of motion and the axis in question, so it is constant. Note that for motion along a line that passes through the axis, the angular momentum is zero.

@RBarryYoung 2015-03-13 00:43:28

"the d there is for point of closest approach between the line of motion" Really? Do you have some kind of reference for that? I ask because I have never heard such a thing before, and it doesn't seem to follow from the canonical case at all. What I would've expected would be that $d$ was the current distance, and rather $\omega$ be adjusted by the cosine of it's angle of incidence wrt the tangent of the current radial circumference (that is, the current angular velocity).

@BowlOfRed 2015-03-13 03:42:22

hyperphysics.phy-astr.gsu.edu/hbase/amom.html $L=mvr\sin(\theta)$. At the point of closest approach, $\sin(\theta)$ is 1 and you have the distance. As the object moves in a straight line, $r\sin(\theta)$ remains constant.

@RBarryYoung 2015-03-27 12:30:00

Could you edit your answer to include some of the information and references from these comments? Thanks.

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